1. ## Riemann integral

http://www.tau.ac.il/~tsirel/Courses...sis3/lect6.pdf

Hi'I need help with the following question:

6g11 page 86

2. ## Re: Riemann integral

Where are you having trouble? I recommend doing problem 6g10 first. Show the solution to that, and we can help you further. (Hint: the answer should be very similar to integrating $x$ from 0 to 1 without the binary expansion).

3. ## Re: Riemann integral

If it helps, here is the idea behind 6g10:

$\beta_k(x) = \begin{cases}1 & \lfloor 2^k x \rfloor \equiv 1\pmod{2} \\ 0 & \text{otherwise}\end{cases}$

In other words, $\beta_k(x) = 1$ when $x \in \bigcup_{n=0}^{2^{k-1}-1} \left[\dfrac{1+2n}{2^k}, \dfrac{n+1}{2^{k-1}}\right)$. Then $\int_{[0,1]} \beta_k(x) = \sum_{n=0}^{2^{k-1}-1} \int_{\left[ \dfrac{1+2n}{2^k},\dfrac{n+1}{2^{k-1}}\right) } 1\cdot dx$ (since these are the only values where the function is 1). Example: $\beta_1(x) =1$ when $x \in \left[\dfrac{1}{2},1\right)$, $\beta_2(x) = 1$ when $x \in \left[\dfrac{1}{4},\dfrac{1}{2}\right)\cup \left[\dfrac{3}{4},1\right)$, etc.

Then $\int_{[0,1]} \sum_{k\ge 1} \dfrac{\beta_k(x)}{2^k} = \sum_{k\ge 1} \dfrac{1}{2^k} \int_{[0,1]} \beta_k(x)$